To do integration by parts:
#uv = intvdu#
I would pick a #u# that gives me an easy time differentiating, and a #dv# that I would easily be able to integrate multiple times if necessary. If the same integral comes back, or variables aren't going away, either it's cyclic or you should switch your #u# and #dv#.
Let:
#u = lnx#
#du = 1/xdx#
#dv = x^(3/2)dx#
#v = 2/5x^(5/2)#
#=> 2/5x^(5/2)lnx - int2/5(x^(5/2)/x)dx#
Nice! That's not too bad.
#= 2/5x^(5/2)lnx - int2/5x^(3/2)dx#
#= 2/5x^(5/2)lnx - 4/25x^(5/2) + C#
#= 10/25x^(5/2)lnx - 4/25x^(5/2) + C#
#= x^(5/2)[10/25lnx - 4/25] + C#
#= 2/25x^(5/2)[5lnx - 2] + C#